The Doppler effect is nothing really new. It’s well-known that Doppler frequency shift occurs for sound as well as light; there are, however, some subtle differences between the two, owing mainly to Einstein’s special theory of relativity. This article describes the classical Doppler effect as it occurs e.g. for sound waves and in which the involved velocities are low enough to be able to neglect relativistic effects. The relativistic version of the Doppler effect as it relates to e.g. electromagnetic waves will be the topic of a later article.

The Doppler effect is described in various levels of detail in a multitude of places [1–3]. What I couldn’t find on the ‘net, though, is a clear mathematical derivation of it by means of plane wave functions in the form

$$\begin{aligned}

E(\mathbf{r},t) & =\frac{1}{2}\Bigl(A_{0}\exp\bigl(i\omega t-i\mathbf{k}\cdot\mathbf{r}\bigr)+\mathrm{c.c.}\Bigr)\\

& =\Re\Bigl\lbrace A_{0}\exp\bigl(i\omega t-i\mathbf{k}\cdot\mathbf{r}\bigr)\Bigr\rbrace\\

& =\Bigl|A_{0}\Bigr|\cos\Bigl(\omega t-\mathbf{k}\cdot\mathbf{r}+\mathrm{arg}\bigl[A_{0}\bigr]\Bigr)\label{plane-wave}

\end{aligned}$$

as I regularly use them here (e.g. in this article) and wherever else they might come in handy. To be as general as possible without being bound to any particular coordinate system, these plane waves should be given in terms of the position vector $\mathbf{r}$ and the reciprocal wave vector $\mathbf{k}$, describing a spatial frequency with direction.

#### Plane Waves

Just about any field distribution can be decomposed into plane waves in order to mathematically propagate them from here to there, as has been done e.g. in the article on the angular spectrum. So if we can describe a linear effect in terms of the plane wave we can describe it for any other wave by decomposing that wave into plane waves, calculating the effect on each of the plane waves and superposing the plane wave solutions to obtain the desired result. Note that this superposition will only work for linear effects / systems.

Now, $E$ in $\eqref{plane-wave}$ can represent the real-valued electric field in the case of light propagation or air pressure in the case of sound waves, $A_{0}$ is the (constant) complex amplitude which contains both magnitude and phase information of the source wave, $\omega$ is the oscillation frequency of the source, $\mathbf{k}$ is the aforementioned wave vector, and “c.c.” stands for the complex conjugate of everything that precedes it. The term

$$A(t)=A_{0}\exp\bigl(i\omega t-i\mathbf{k}\cdot\mathbf{r}\bigr)\label{plane-wave-complex}$$

is called the *phasor* and fully describes the wave. We will use this phasor a lot for linear systems, because it allows us to work with the relatively simple exponential functions instead of the more complicated cosine. Both are equally valid, though.

The frequency $\omega^\prime$ that the receiver at position $\mathbf{r}$ detects is simply the time derivative of the phase of the phasor $A(t)$:

$$\begin{aligned}

\omega^\prime & =\partial_{t}\arg\bigl[A(\mathbf{r},t)\bigr] \label{frequency}\\

& =\omega

\end{aligned}$$

when the source amplitude $A_{0}$ is constant. As long as nothing moves, the receiver sees (or hears) the same frequency that was sent. When things get moving, however, this changes as a result of the Doppler effect.

There are actually two slightly different Doppler effects in the classical domain. When we have a transporting medium like we have with sound waves, it makes a difference whether the wave source is stationary with respect to the medium (and only the receiver moves) or whether the source is moving, because the velocity $c$ of the waves is defined relative to the medium. Both cases can also be combined in various ways.

#### Moving Receiver

The mathematically most straightforward case is the moving receiver with a stationary source: The vector $\mathbf{r}$ in $\eqref{plane-wave}$ describes the position of the receiver in space (the source is assumed to be at the origin of the coordinate system). The motion of the receiver is described by

$$\mathbf{r}=\mathbf{r}_{0}+\mathbf{v}_{R}t=\mathbf{r}_{0}+v_{R}\mathbf{\hat{v}}_{R}t$$

in which $v_{R}=\bigl|\mathbf{v}_{R}\bigr|$ is its velocity, $\mathbf{\hat{v}}_{R}$ is a unit vector in the direction of motion and $\mathbf{r}_{0}$ is its initial position. For the moving receiver, $\eqref{plane-wave}$ becomes

$$A_{R}(\mathbf{r},t)=A_{0}\exp\Bigl[i\omega t-i\mathbf{k}\cdot\bigl(\mathbf{r}_{0}+\mathbf{v}_{R}t\bigr)\Bigr]\label{plane-wave-R}$$

The received frequency is determined analogous to $\eqref{frequency}$:

$$\begin{aligned}

\omega_{R}^\prime & =\partial_{t}\arg\bigl[A_{R}(\mathbf{r},t)\bigr]\\

& =\omega-\mathbf{k}\cdot\mathbf{v}_{R}\\

& =\omega\biggl(1-\frac{\mathbf{k}\cdot\mathbf{v}_{R}}{\omega}\biggr)\\

& =\omega\biggl(1-\frac{\mathbf{\hat{k}}\cdot\mathbf{v}_{R}}{c}\biggr)

\end{aligned}\label{frequency-R}$$

where we used

$$\mathbf{k}=\mathbf{\hat{k}}\frac{\omega}{c}\label{normalized-wave-vector}$$

Hence, there is a frequency shift which depends on the direction of the receiver motion relative to the direction of the wave coming from the source. If the receiver approaches the source directly ($\mathbf{k}$ is antiparallel to $\mathbf{v}$), the frequency increase is maximum. The observed frequency shift due solely to the change of the received phase as a result of the receiver motion within the field. Neither the spatial distribution nor the oscillation frequency of the field has changed.

Also note that the case $\omega\rightarrow\infty$ (describing a sonic boom) is not achievable since the receiver velocity cannot reach infinity.

#### Moving Source

Things change a bit with a source on the move. We could try and switch our frame of reference to the moving source to re-create the situation with a moving receiver and a stationary source in the new coordinates, apply $\eqref{plane-wave-R}$ and $\eqref{frequency-R}$, and be done with it. In this case we would disregard, however, that the medium is now also moving with respect to our coordinate system. And since the wave velocity $c$ is defined relative to the medium, our result would be wrong.

To correct this error, we need to figure out the new wave vector $\mathbf{k}_{S}$ in the new frame of reference in which the medium is moving. We’ll start by decomposing the medium motion into a motion parallel to the initial wave vector $\mathbf{k}$ and one orthogonal to it. The magnitude of the former is

$$\Delta c=-\mathbf{\hat{k}}\cdot\mathbf{v}_{S}=-\frac{c}{\omega}\mathbf{k}\cdot\mathbf{v}_{S}$$

in which $\mathbf{v}_{S}$ is the velocity vector of the source in the reference frame in which the medium is at rest and hence $-\mathbf{v}_{S}$ is the velocity vector of the medium in the moving frame. This component of the medium velocity directly alters the phase velocity $c$ of the wave since the phase velocity is defined relative to it. The other component, orthogonal to $\mathbf{k}$ and parallel to the wave fronts, will shift the wave sideways, but since the plane wave is infinitely extended in that direction, such a shift causes no observable change and can thus be neglected. Since the only change is in the direction $\mathbf{\hat{k}}$ of the wave vector, that direction does not change, merely its magnitude is affected.

Then, with the modified phase velocity

$$c^\prime=c+\Delta c=c\biggl(1-\frac{\mathbf{k}\cdot\mathbf{v}_{S}}{\omega}\biggr)\label{velocity-S}$$

we have the new wave vector

$$\begin{aligned}

\mathbf{k}_{S} & =\frac{\omega}{c^\prime}\mathbf{\hat{k}}\\

& =\mathbf{k}\biggl(1-\frac{\mathbf{\hat{k}}\cdot\mathbf{v}_{S}}{c}\biggr)^{-1}\\

& =\mathbf{k}\biggl(1-\frac{\mathbf{k}\cdot\mathbf{v}_{S}}{\omega}\biggr)^{-1}

\end{aligned}\label{wave-vector-S}$$

As our frame of reference transformation is purely translational, the same wave vector $\mathbf{k}_{S}$ is observed in the moving as well as the fixed frame. Note that a simple addition of velocities as in $\eqref{velocity-S}$ is only possible in classical physics. To determine the received frequency as before, we’ll remain in the moving frame, noting that in this frame the receiver also moves with $-\mathbf{v}_{S}$ as does the medium. We insert $\eqref{wave-vector-S}$ into $\eqref{plane-wave-complex}$,

$$A_{S}(\mathbf{r},t)=A_{0}\exp\Bigl[i\omega t-i\mathbf{k}_{S}\cdot\bigl(\mathbf{r}_{0}-\mathbf{v}_{S}t\bigr)\Bigr]$$

and determine $\omega_{S}^\prime$ analogous to $\eqref{frequency}$:

$$\begin{aligned}

\omega_{S}^\prime & =\partial_{t}\arg\bigl[A_{S}(\mathbf{r},t)\bigr]\\

& =\omega+\mathbf{k}\cdot\mathbf{v}_{S}\biggl(1-\frac{\mathbf{k}\cdot\mathbf{v}_{S}}{\omega}\biggr)^{-1}\\

& =\omega\biggl(1+\frac{\mathbf{k}\cdot\mathbf{v}_{S}}{\omega-\mathbf{k}\cdot\mathbf{v}_{S}}\biggr)\\

& =\omega\biggl(1-\frac{\mathbf{k}\cdot\mathbf{v}_{S}}{\omega}\biggr)^{-1}\\

& =\omega\biggl(1-\frac{\mathbf{\hat{k}}\cdot\mathbf{v}_{S}}{c}\biggr)^{-1}\label{frequency-S}

\end{aligned}$$

Note that if the component of the source velocity parallel to the wave vector equals the wave velocity $c$ in the medium, both $\mathbf{k}$ and $\omega_{S}^\prime$ become infinite, resulting in a sonic boom in the case of sound waves. Also, the apparent frequency shift is now due the receiver motion within the wave field as above, as well as the change of the wavelength within the wave field. The frequency of the wave again remains the same.

#### Both Moving

When both source and receiver are moving, we only need to alter $\eqref{plane-wave-S}$ to additionally account for the receiver motion:

$$A_{RS}(\mathbf{r},t)=A_{0}\exp\Bigl[i\omega t-i\mathbf{k}_{S}\cdot\bigl(\mathbf{r}_{0}-\mathbf{v}_{S}t+\mathbf{v}_{R}t\bigr)\Bigr]$$

resulting in a received frequency of

$$\begin{aligned}

\omega_{RS}^\prime & =\omega\biggl(1-\frac{\mathbf{k}\cdot\mathbf{v}_{R}}{\omega}\biggr)\biggl(1-\frac{\mathbf{k}\cdot\mathbf{v}_{S}}{\omega}\biggr)^{-1}\\

& =\omega\biggl(1-\frac{\mathbf{\hat{k}}\cdot\mathbf{v}_{R}}{c}\biggr)\biggl(1-\frac{\mathbf{\hat{k}}\cdot\mathbf{v}_{S}}{c}\biggr)^{-1}

\end{aligned}$$

The correction factor to the frequency $\omega$ in this case is the product of the correction factors in $\eqref{frequency-R}$ and $\eqref{frequency-S}$. Also, if both source and receiver move with the same velocity relative to the medium, the received frequency is again equal to the source frequency $\omega$.

Note that the frequency $\omega_{RS}^\prime$ depends on the absolute velocities of source and receiver, not just how they move relative to each other. This is due to the crucial role of the medium in classical physics.

#### Echo from a Moving Reflector

A stationary source emits a wave which is reflected back towards the source by a moving receiver. A Doppler shift as above occurs in both directions. The frequency observed by the source can again be found by a combination of the first two cases. The outgoing wave is received by the reflector with an observed frequency of $\eqref{frequency-R}$,

$$\omega_{SRS}^\prime=\omega\biggl(1-\frac{\mathbf{\hat{k}}\cdot\mathbf{v}_{R}}{c}\biggr)$$

The reflector in turn becomes a moving source which emits waves with the modified frequency $\omega_{SRS}^\prime$ which is observed by it and also a modified wave vector corresponding to that frequency

$$\mathbf{k}^\prime=-\mathbf{k}\biggl(1-\frac{\mathbf{\hat{k}}\cdot\mathbf{v}_{R}}{c}\biggr)$$

cf. also $\eqref{normalized-wave-vector}$. The minus sign results from the change of direction upon reflection. The frequency observed by the source upon reception of the waves emitted by the reflector is given by $\eqref{frequency-S}$ in which $\mathbf{v}_{R}=\mathbf{v}_{S}$ (both describing the motion of the reflector) and $\omega$ and $\mathbf{k}$ have been replaced by $\omega_{SRS}^\prime$ and $\mathbf{k}^\prime$, respectively:

$$\begin{aligned}

\omega_{SRS}^{\prime\prime} & =\omega_{SRS}^\prime\biggl(1-\frac{\mathbf{k}^\prime\cdot\mathbf{v}_{R}}{\omega_{SRS}^\prime}\biggr)^{-1}\\

& =\omega\biggl(1-\frac{\mathbf{k}\cdot\mathbf{v}_{R}}{\omega}\biggr)\biggl(1+\frac{\mathbf{k}\cdot\mathbf{v}_{R}}{\omega}\biggr)^{-1}\\

& =\omega\biggl(1-\frac{\mathbf{\hat{k}}\cdot\mathbf{v}_{R}}{c}\biggr)\biggl(1+\frac{\mathbf{\hat{k}}\cdot\mathbf{v}_{R}}{c}\biggr)^{-1}\\

& =\omega\frac{c-\mathbf{\hat{k}}\cdot\mathbf{v}_{R}}{c+\mathbf{\hat{k}}\cdot\mathbf{v}_{R}}

\end{aligned}

\label{frequency-SRS}$$

#### Decomposition

Note that the perceived Doppler shift in $\eqref{frequency-SRS}$ depends solely on the component of the reflector motion which is parallel to the propagation velocity of the plane wave. If the source is not a source of plane waves, which it cannot be for plane waves are infinitely extended, we can decompose whatever waves the source does emit into an angular spectrum of plane waves, as mentioned above. See also the article on the angular spectrum in the context of Fourier diffraction. Each of these plane waves has a different direction $\mathbf{\hat{k}}$ and thus will generally cause a different detected frequency shift upon reflection and subsequent reception. The source will thus see (or hear) a whole spectrum of reflected frequencies.

This is compounded by the finite extent of the receiver and thus a finite cross-section for reflection. Each incoming plane wave is spatially filtered upon reflection so that the reflected portion of it must again be decomposed into its angular spectrum, each component of which is received with a different Doppler shift back at the source. The reflection spectrum can possibly be deconvoluted and analyzed to reveal information about the reflection cross-section, but such an analysis exceeds the scope of this article.

[1] Doppler effect [Wikipedia]

[2] Doppler effect [James B. Calvert]

[3] Doppler Shift for Sound and Light [Reflections on Relativity]

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